A boy of mass 40 kg is hanging from the horizontal branch of a tree. The tension in his arms is minimum when the angle between the arms is:-

Options

A. `0^(@)`
B. `90^(@)`
C. `120^(@)`
D. `180^(@)`

Explanation

To solve the problem of finding the angle between the arms of a boy hanging from a tree branch at which the tension in his arms is minimized, we can follow these steps:

### Step 1: Understand the Forces Acting on the Boy
The boy has a mass of 40 kg, and thus the gravitational force acting on him (weight) is given by:
\[ F_g = mg = 40 \, \text{kg} \times g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 2: Set Up the Tension Forces
When the boy hangs from the branch, he exerts tension in both arms. Let \( T \) be the tension in each arm. The angle between the arms is denoted as \( \theta \). The vertical components of the tension must balance the weight of the boy.

### Step 3: Resolve the Tension into Components
The vertical component of the tension can be expressed as:
\[ T_y = T \cos(\frac{\theta}{2}) \]
Since there are two arms, the total upward force due to tension is:
\[ 2T \cos(\frac{\theta}{2}) \]

### Step 4: Set Up the Equation for Equilibrium
For the boy to be in equilibrium, the total upward force must equal the downward gravitational force:
\[ 2T \cos(\frac{\theta}{2}) = mg \]
Substituting \( mg \) with \( 40g \):
\[ 2T \cos(\frac{\theta}{2}) = 40g \]

### Step 5: Solve for Tension
From the equation, we can express \( T \) as:
\[ T = \frac{40g}{2 \cos(\frac{\theta}{2})} = \frac{20g}{\cos(\frac{\theta}{2})} \]

### Step 6: Minimize the Tension
To minimize the tension \( T \), we need to maximize the denominator \( \cos(\frac{\theta}{2}) \). The maximum value of \( \cos(x) \) occurs when \( x = 0 \), which means:
\[ \frac{\theta}{2} = 0 \implies \theta = 0 \]

### Conclusion
Thus, the angle between the arms when the tension is minimized is:
\[ \theta = 0^\circ \]